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We know how to find the area of an equilateral triangle
What I want to do in this video is attempt to find the area of a a
I know I'm mispronouncing it Koch snowflake
And the way you construct one
is you start one with an equilateral triangle
And on each of the sides you split them into thirds
And then the middle third you put
another smaller equilateral triangle
And that's after one pass
And on the next pass you do that for all of the sides here,
So little one here here here here here here here
I think you get the general idea that's a nice pass
So the next pass you do it for all these sides
And what's really neat about this and
we showed this on the previous video is that
you have a figure here that has an infinite perimeter
but we're about to see in this video it actually has a finite area
Which is kinda of interesting to think about
So let's start with a clean equilateral triangle right over here
We're gonna assume that each of the sides have length S,
So, it's going to be a clean equilateral triangle each of the sides
I'll draw that a little bit neater, each of the sides have length S
And so, what I wanna do that is keep to track of two things
I wanna keep track of the sides of this triangle
as we are or as it as it turns into a snowflake
I gonna keep track of the number sides
and I gonna keep track of the area
after each pass of adding more smaller triangles
So this is gonna be our count of the area
Actually, let me give myself a little bit more real estate
because I have a feeling I might need to use it
So might keep this is the sides
I'm gonna write up here
and then this is our running count of the area down here
So right when we start,
We have three sides
And our area we are to figure out in the previous video is going to be
we assume that each of the sides are length S
is going to be square root of three square root
of three S squared over four
Fair enough that's just a simple equilateral triangle
Now we're going to take each of the sides divide them into thirds
So we're gonna take each sides divid it into thirds
And then, in that middle third we're going
to add another smaller equilateral triangle
So it will look like that on that side right over there
I want you to think about what we're doing each side right here
So before I did this, this is just one side
Then I split it into thirds and that middle third
I put I said you put two sides in there
I put an equilateral triangle
So one side has now turned into one two three four sides
So every time we do a pass
of the making the snowflake more intricate
Each side will turn into four sides
So you can imagine if we do this on all three sides,
we have four times three which is now 12 sides
So if you multiply these times four,
so times four this gets us to 12 sides now
We can count them out just
to make sure we're on logic this correct,
One, two, three, four, five, six, seven,
Eight, nine, ten, eleven, twelve sides
And now what is the area now?
What's going to be the area
of which of all yellow equilateral triangles
plus the area of each of these smaller ones?
And what's the area of each of these smaller ones?
Well, first of all we have three of them
There's three of these each of these smaller ones
And then we use the formula for an equilateral triangle again
So it's going to be the square root of three times S squared
For now the length of each of these sides
Each these sides are equal to these smaller equilateral triangles
They aren't S anymore
they are S over three
remember this length right over here is S over three
so this is going to be S over three as well
Every pass,
the sides of the equilateral triangle
become one third of the previous pass
So this is not going to be S squared anymore
It's gonna be S over three squared
and then all of that over four
Then let's do another pass
So, I'm gonna add these triangles right over here
Then add these right over there
And this is the last pass where I actually attempt
to draw all of the triangles over there
So how many how many sides am I going to have,
first of all, after I do another pass?
Well the previous pass I have 12 sides,
each of those 12 sides are now going turn into four new sides
When I add these little orange bumps there
So when I multiply it times four again
'm gonna multiply it times four,
so now I'm going to have 48 sides
I'm gonna have 48 sides
And how many new triangles,
so what's the area what's gonna be the yellow area
plus the blue area plus the orange area
So how many new orange triangles do I have?
Well I'm adding a new orange triangle
to each of the sides for the previous pass
And the previous pass I have 12 sides,
so now I'm going to add 12 orange triangles
I'm going to add 12 orange triangles
And actually let me write that
or I'll just write 12 orange triangles but it's really,
I just multiply it times four
And then I'm going to have times the square root of three
And now this is not going to be S over three anymore
This is going to be s over nine
These have one third dimensions of these blue triangles
So this is going to be S over nine squared,
S over nine squared over four
And so, I think you might start to see the pattern building
if we do another pass after this one
Move to the right a little bit what will that look like
Let gonna do this in a different color that I haven't used yet
Let me see if haven't use this pink yet
So now we're going to have,
I'm going to have the previous number sides
that's my number of new triangles
48 times the square root of three times S over --
I'm going to, now these are gonna be one third of these
S over 27 to the second power all of that over four
And I'm going to keep adding an infinite number of terms of this
to get the area of a true Koch snowflake
So I'm just gonna keep doing this over and over again
So that trick really is is finding this infinite sum
and see if we get a finite number over here
So the first things I wanna do just to simplify,
Well let me just rewrite it a little bit?
Let me rewrite it a little bit different over here
So the first thing that's kind of it, that might be obvious is that
We can throughout the square root of three S squared over four
So let me just factor that out
So if we factor a square root of three S squared over four
from all of the terms then this term right
over here will become a one
This term right over here is going to become a three,
let see we factor a squared of three we factor out of four
and we factored out the S squared
We factored out all in the S squared,
so now it's going to have plus three times
one third three times one third squared
That's all we have left here,
We have the one third squared
and then we have this three
And I'm not simplifying this on purpose
so that we see a pattern emerge
And then this next term right over here plus
so this 12 is still going to be there
but I'm gonna write that as three times four
And we're gonna lose, this that we're factoring out
the square root of three
we're factoring out the four
we're factoring out the S squared
And so we're gonna be left with three squared
that's what this is down here, squared
So this is one over one third squared and then that squared
So that's what we're left with that orange term
And then we're going to the pink term
This pink term, this is 48 it is just three times four times four
Three times four
I'll write four squared here
cause each time we're gonna multiply times four again
So next is gonna be four to the third
Because we're really each iteration,
each sides turns into four sides
that's where that came from
Four squared we lose we're factoring out the squared of three
we're factoring out the four we're, factoring out the S squared
And all we're left is one over three to the third power squared
So, times one over three to the third power squared
and we'll just gonna keep going like that forever
Keep going like that forever
So, on each each step we're incrementing,
we're multiply by four and we're also multiplying
we're, I guess we say, the power of this four as incrementing
So it goes from there's usually fourth of the zeroth power here
We have a one here
You can kind of imagine implicitly the fourth
of the first power four squared and then it'll go four to the third
And then we have this power is also increment
Three to the first, three to the second, three to the third
We see that this power is always one more than that
And it'll be much easier to calculate this infinite
which is gonna turn into infinite geometric series
if those were actually the same power
So, what I wanna do is I wanna
increase the power four in all of those
But I can't just willy-nilly multiply everything by four
If I'm gonna multiply everything by four
I also need to divide everything by four
So, what I'm gonna do in this right over here
is I'm gonna multiply and divide everything by four
So, if we divide by four I can do that on the outside
So, I'm gonna multiply it one fourth times this right over here
And, so I'm dividing by four and here
I'm going to multiply this by four
So, I'm not going to be changing the value of the actual thing
So, this is going to be four plus three times four
plus three times four squared four to the third
And so what was cool about this is now
at the power of power and the power of this three down here
are going to be the same power
But it still seems a little weird because we're
taking this one over three squared and we're squaring it,
one over three to the third then we're squaring it
And here we just have to realize,
so this is always gonna be squared
and this is the thing that is incrementing
But in general,
if I have one over three to the end
and I'm squaring it
This is equal to one over three to the 2N power,
which is, so I'm just multiplying it by two, right?
If I'm raising something to an exponent
then raising that to an exponent,
that's just multiplying it times the or raising
to the or raising to the Nth times to exponent
And this is the exact same thing
as one over three squared raised to the Nth power
So we could actually switch these two exponents
in a very legitimate way
And then let me rewrite everything
'cause I do wanna do too much on this one step right over here
So, part right over her gives us square of three S squared
over 16 and then that's going to be times,I'll open
and close parenthesis
So then we have four plus,
then in blue,
I'll write three times four to the first power
And then I can write this,
I can rewrite this as one times one third
We could view this as one third squared
or we could view this, we could view this
as one over three to the first power square
or we can view this as one over three squared to the first power
And I'm gonna write it that way
So time one ninth to the first power
And then plus three times four squared
And then these we can write as times one ninth to the second power
And then these where we can write plus three times four to the third
times
and this is we could write
this is on over 27 to the second power
But we could also write this based on what we saw over here
Let me make this clear
One over three to the third to the second power
This is the same thing as one over three squared to the third power
That's what we showed right over here
So, this is equivalent to one ninth to the third power
Now we start to see the pattern
and starting to clean up a little bit
And let me just do one more step,
and then we'll finish this to the next video
So this is equal to square root of three S squared
over 16 times four plus four plus three times,
this is four ninths plus,
the next term is three times four ninths squared
And then we have plus times four over nine to the third power
And we're just gonna keep on and on and on and on
taking three time four ninths
To the successively larger and larger power
So, this is what we have to find the sum of
to find our area and we're gonna do that on the next video
We're gonna use some of the tools
we've used to find the sums of infinite geometric series
But we're kind of going to redo it in the next video
just so you don't have to remember that formula or that proof